Pretty much this, this kind of question is a great example of why rigour is important in mathematics. I have been out all day so I STILL haven't looked at this properly, but will do so tomorrow and post whatever I can come up with.
if your proof is anything like mind, a post online isn't realistic. Even if i was computer savvy it would require too many diagrams and explanations. The approximate process is the following:
1. do what realisenothing did and show that they cannot be arbitrarily close
2. construct a diagram with the two individuals who get shot (say points X and Y in the plane). divide the plane up into areas defined by the following lines:
a) the perpendicular bisector of XY
b) the line through Y perpendicular to XY
c) the line through X perpendicular to XY
(call the area of the plane between lines b) and c) area B, and call the area outside these lines A and A' (effectively the same due to symmetry))
3. from this diagram you can prove the impossibility of the following cases geometrically:
a) all 8 points in A or A' , but none in B
b) 1 point in B, and 7 in A or A' (this is basically the same proof as a) - it is a blatant corollary. I probably shouldnt consider it a separate proof)
c) 2 points in B, and 6 in A or A'
d) 3+ points in B, and 5- points in A or A'
also make sure you consider border cases as well.
for each case you should reach a definitive mathematical/geometrical contradiction/impossibility. don't just settle for: "well it is visually apparent that this particular construct won't work"
have fun
